How do you integrate #x^3/sqrt(144-x^2)#?

1 Answer
Jim H
Aug 23, 2015

Use substitution with #u = 144-x^2#.

Explanation:

#int x^3/sqrt(144-x^2) dx#

Let #u = 144-x^2#, so that #du = -2xdx# and #x^2 = 144-u#

#int x^3/sqrt(144-x^2) dx = int x^2/sqrt(144-x^2)" " xdx#

# = -1/2 int x^2/sqrt(144-x^2)" "(-2 x)dx#

# = -1/2 int (144- u)/sqrtu" "du#

# = -1/2 int (144/sqrtu - sqrtu)" "du#

# = -1/2 int (144u^(-1/2) -u^(1/2))" "du#

# = -1/2[288u^(1/2)-2/3u^(3/2)] +C#

# = -144(144-x^2)^(1/2)+ 1/3(144-x^2)^(3/2) +C#

# = -144sqrt(144-x^2)+ 1/3(sqrt(144-x^2))^3 +C#

Related questions
See all questions in Integration by Trigonometric Substitution
Impact of this question
1983 views around the world
You can reuse this answer
Creative Commons License